The ground state magnetic property of B_{2} a | vedclass

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(A) According to molecular orbital theory, the electronic configuration for $B_{2}$ ($10$ electrons) is: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^

Vedclass · Accepted Answer

$B_{2}$ paramagnetic and $C_{2}$ diamagnetic

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(A) According to molecular orbital theory, the electronic configuration for $B_{2}$ ($10$ electrons) is: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{x}^{1} = \pi 2p_{y}^{1}$.Since it contains two unpaired electrons, $B_{2}$ is paramagnetic.The electronic configuration for $C_{2}$ ($12$ electrons) is: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}$.Since all electrons are paired, $C_{2}$ is diamagnetic.

The ground state magnetic property of $B_{2}$ and $C_{2}$ molecules will be

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